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The Total Symmetry of Three-dimensional Crystals

Part VI
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e-mail :

Sequel to Group Theory

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We'll start with reminding the reader about the "Important Remark" near the end of Part III, a Remark concerning the direction of reading products of group elements, like, say, apq. We read such products (from that Remark onwards) from back to front. Thus (with respect to apq) first  q, then  p, and then  a.
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Sequel to Cyclic Groups of composite order

In the previous document we considered the cyclic group C12 in its realization of the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} under addition modulo 12. The initial version of that realization was such that the elements were arranged in there natural order. See the previous document Table 5.1. Then we rearranged the elements such that the subgroup H = {0, 6} stood out. We discovered that this subgroup supported a homomorphism :  The set of cosets of this subgroup (including the subgroup itself) was itself a group. This group then was accordingly the image group of the homomorphism associated with this subgroup. Moreover this subgroup was mapped by the homomorphism onto the identity element of the image group. It was called the kernel of that homomorphism.

We now consider yet another rearrangement of the elements of the group, such that now the subgroup P = {0, 4, 8} will stand out. Indeed, when we try to generate the whole group with the element 4, namely by a repeated application of it we find :
4 + 4 (mod 12) = 8, i.e. now element 8 is generated.
4 + 4 + 4 (mod 12) = 0, i.e. now the element 0 is generated.
4 + 4 + 4 + 4 (mod 12) = 4, i.e. we are back at element 4.
When we try to generate the whole group with the element 8, we get :
8 + 8 (mod 12) = 4.
8 + 8 + 8 (mod 12) = 0.
8 + 8 + 8 + 8 (mod 12) = 8.
So we can only generate the elements 0, 4, and 8. We stay within the subset {0, 4, 8}, and thus this subset is closed under the group operation addition modulo 12. It contains the identity element. And because
0 + 0 (mod 12) = 0 (where 0 is element of the subset),
4 + 8 (mod 12) = 0 (where 4, 8 and 0 are elements of the subset), and
8 + 4 (mod 12) = 0 (where 8, 4 and 0 are elements of the subset),
every element of the subset {0, 4, 8} has an inverse.
Associativity is guaranteed because we still have to do with addition modulo 12 which is associative. So now we are sure that the subset P = {0, 4, 8} is a subgroup.

The group table of our group C12 in its realization in finite arithmetic, and with its elements rearranged such that the subgroup (0, 4, 8} and its cosets stands out, is as follows :

 + mod 12 0 4 8 1 5 9 2 6 10 3 7 11 period 0 0 4 8 1 5 9 2 6 10 3 7 11 1 4 4 8 0 5 9 1 6 10 2 7 11 3 3 8 8 0 4 9 1 5 10 2 6 11 3 7 3 1 1 5 9 2 6 10 3 7 11 4 8 0 12 5 5 9 1 6 10 2 7 11 3 8 0 4 12 9 9 1 5 10 2 6 11 3 7 0 4 8 4 2 2 6 10 3 7 11 4 8 0 5 9 1 6 6 6 10 2 7 11 3 8 0 4 9 1 5 2 10 10 2 6 11 3 7 0 4 8 1 5 9 6 3 3 7 11 4 8 0 5 9 1 6 10 2 4 7 7 11 3 8 0 4 9 1 5 10 2 6 12 11 11 3 7 0 4 8 1 5 9 2 6 10 12

Table 6.1

Table of C12   (version c).
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P = {0, 4, 8} is a subgroup. We will determine its cosets.
Its left coset by the element 0 is 0P = 0{0, 4, 8} = {0 + 0 (mod 12), 0 + 4 (mod 12), 0 + 8 (mod 12)} = {0, 4, 8}.
Its right coset by the element 0 is P0 = {0 + 0 (mod 12), 4 + 0 (mod 12), 8 + 0 (mod 12)} = {0, 4, 8}.  So we can say that
0P = P0 = {0, 4, 8} = P.

In the same way we can determine all cosets of this subgroup P, and because the operation addition modulo 12 is commutative, we know that all left and right cosets of P with respect to a particular element of the group are equal, which means that the subgroup P = {0, 4, 8} is a normal subgroup in G (which is the whole group, namely our group C12, realized in modulo 12 arithmetic) :

4P = 4{0, 4, 8} = {4 + 0 (mod 12), 4 + 4 (mod 12), 4 + 8 (mod 12)} = {4, 8, 0} = P (because, of course, {4, 8, 0} = {0, 4, 8}). So
4P = P4 = P = {0, 4, 8}.

8P = 8{0, 4, 8) = {8 + 0 (mod 12), 8 + 4 (mod 12), 8 + 8 (mod 12)} = {8, 0, 4} = P. So
8P = P8 = P = {0, 4, 8}.

So the cosets of the subgroup P by its own elements are all equal to P itself.

1P = 1{0, 4, 8} = {1 + 0 (mod 12), 1 + 4 (mod 12), 1 + 8 (mod 12)} = {1, 5, 9}. So
1P = P1 = {1, 5, 9}.

5P = 5{0, 4, 8} = {5 + 0 (mod 12), 5 + 4 (mod 12), 5 + 8 (mod 12)} = {5, 9, 1} = 1P. So
5P = P5 = 1P = P1 = {1, 5, 9}.

9P = 9{0, 4, 8} = {9 + 0 (mod 12), 9 + 4 (mod 12), 9 + 8 (mod 12)} = {9, 1, 5} = 5P. So
9P = P9 = 5P = P5 = P1 = 1P = {1, 5, 9}.

2P = P2 = 2{0, 4, 8} = {2 + 0 (mod 12), 2 + 4 (mod 12), 2 + 8 (mod 12)} = {2, 6, 10}. So
2P = P2 = {2, 6, 10}.
6P = P6 = 6{0, 4, 8} = {6 + 0 (mod 12), 6 + 4 (mod 12), 6 + 8 (mod 12)} = {6, 10, 2} = P2. So
6P = P6 = P2 = 2P = {2, 6, 10}.
10P = 10{0, 4, 8} = {10 + 0 (mod 12), 10 + 4 (mod 12), 10 + 8 (mod 12)} = {10, 2, 6} = 6P. So
10P = P10 = 6P = P6 = P2 = 2P = {2, 6, 10}.

3P = 3{0, 4, 8} = {3 + 0 mod 12), 3 + 4 (mod 12), 3 + 8 (mod 12)} = {3, 7, 11}. So
3P = P3 = {3, 7, 11}.
7P = 7{0, 4, 8} = {7 + 0 (mod 12), 7 + 4 (mod 12), 7 + 8 (mod 12)} = {7, 11, 3} = 3P. So
7P = P7 = 3P = P3 = {3, 7, 11}.
11P = 11{0, 4, 8} = {11 + 0 (mod 12), 11 + 4 (mod 12), 11 + 8 (mod 12)} = {11, 3, 7} = 7P. So
11P = P11 = 7P = P7 = 3P = P3 = {3, 7, 11}

The cosets {0, 4, 8}, {1, 5, 9}, {2, 6, 10} and {3, 7, 11} can be written as :
{0, 0 + 4, 0 + 4 + 4},
{1, 1 + 4, 1 + 4 + 4},
{2, 2 + 4, 2 + 4 + 4}, and
{3, 3 + 4, 3 + 4 + 4}.

So we can represent these sets by the numbers 0, 1, 2, and 3, which now themselves form a set C = {0, 1, 2, 3}.
This means that we can set up a second homomorphism -- The first homomorphism was set up in the previous document with respect to the subset {0, 6} -- which is a homomorphism with respect to the subgroup {0, 4, 8}, which maps
the subset {0, 4, 8} of our group G onto the element  0  of C,
the subset {1, 5, 9} of our group G onto the element  1  of C,
the subset {2, 6, 10} of our group G onto the element  2  of C, and
the subset {3, 7, 11} of our group G onto the element  3  of C.

If we now replace the subsets (cosets) {0, 4, 8}, {1, 5, 9}, {2, 6, 10} and {3, 7, 11} in the above group table (Table 6.1) by those numbers 0, 1, 2 and 3 , we see that the structure which emerges is that of the arithmetic of addition modulo 4, i.e. a realization of the cyclic group of order 4,  C4  :

 + mod 4 0 1 2 3 period 0 0 1 2 3 1 1 1 2 3 0 4 2 2 3 0 1 2 3 3 0 1 2 4

Table 6.2

Table of C4.
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So the group C4, which in Table 6.2 is realized as addition modulo 4, is the homomorphic image of the group C12 with respect to the latter's subgroup {0, 4, 8}.
We can see that the global structure of C12 as it appears in version c (Table 6.1) reappears in its homomorphic image with respect to the subgroup {0, 4, 8}. In order to make this clear we reproduce the corresponding group tables once again for easy comparison. The homomorphism is called  j  :

 + mod 12 0 4 8 1 5 9 2 6 10 3 7 11 0 0 4 8 1 5 9 2 6 10 3 7 11 4 4 8 0 5 9 1 6 10 2 7 11 3 8 8 0 4 9 1 5 10 2 6 11 3 7 1 1 5 9 2 6 10 3 7 11 4 8 0 5 5 9 1 6 10 2 7 11 3 8 0 4 9 9 1 5 10 2 6 11 3 7 0 4 8 2 2 6 10 3 7 11 4 8 0 5 9 1 6 6 10 2 7 11 3 8 0 4 9 1 5 10 10 2 6 11 3 7 0 4 8 1 5 9 3 3 7 11 4 8 0 5 9 1 6 10 2 7 7 11 3 8 0 4 9 1 5 10 2 6 11 11 3 7 0 4 8 1 5 9 2 6 10
 + mod 4 0 1 2 3 0 0 1 2 3 1 1 2 3 0 2 2 3 0 1 3 3 0 1 2
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The group C12 has still more subgroups that support a homomorphism to be formed, namely (conform the notation of our tables) the subgroups {0, 3, 6, 9} and {0, 2, 4, 6, 8, 10}.
We'll begin with the former, namey {0, 3, 6, 9} which we will call  R.

 + mod 12 0 3 6 9 1 4 7 10 2 5 8 11 period 0 0 3 6 9 1 4 7 10 2 5 8 11 1 3 3 6 9 0 4 7 10 1 5 8 11 2 4 6 6 9 0 3 7 10 1 4 8 11 2 5 2 9 9 0 3 6 10 1 4 7 11 2 5 8 4 1 1 4 7 10 2 5 8 11 3 6 9 0 12 4 4 7 10 1 5 8 11 2 6 9 0 3 3 7 7 10 1 4 8 11 2 5 9 0 3 6 12 10 10 1 4 7 11 2 5 8 0 3 6 9 6 2 2 5 8 11 3 6 9 0 4 7 10 1 6 5 5 8 11 2 6 9 0 3 7 10 1 4 12 8 8 11 2 5 9 0 3 6 10 1 4 7 3 11 11 2 5 8 0 3 6 9 1 4 7 10 12

Table 6.3

Table of C12   (version d).
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The cosets of the subset R = {0, 3, 6, 9} are the sets
{0, 3, 6, 9}, {1, 4, 7, 10}, and {2, 5, 8, 11}. Like we did earlier it can be seen from the group table (version d, as depicted right above) that the left and right cosets of R are equal for every paricular element, i.e. xR = Rx, in which  x  is any element of the group (i.e. our group C12, called G, and realized as {0, 1, 2, 3, ..., 11}, + mod 12). This implies that the subset R = {0, 3, 6, 9} is a normal subgroup in G.

These cosets can be renamed in the following way :

{0, 3, 6, 9} -- {0, 0 + 3, 0 + 3 + 3, 0 + 3 + 3 + 3} -- 0.
{1, 4, 7, 10} -- {1, 1 + 3, 1 + 3 + 3, 1 + 3 + 3 + 3} -- 1.
(2, 5, 8, 11} -- {2, 2 + 3, 2 + 3 + 3, 2 + 3 + 3 + 3} -- 2.

These numbers, representing the cosets of R form a set D = {0, 1, 2}.
So now we can set a homomorphism of our group G, with respect to the subgroup R = {0, 3, 6, 9}, which maps
the subset (coset) {0, 3, 6, 9} of our group G onto the element  0  of the set D,
the subset (coset) {1, 4, 7, 10} of our group G onto the element  1  of the set D, and
the subset (coset) {2, 5, 8, 11} of our group G onto the element  2  of the set D.

Let's call this homomorphism  m.

If we now replace the subsets (cosets) {0, 3, 6, 9}, {1, 4, 7, 10} and {2, 5, 8, 11} in the above group table (Table 6.3) by those numbers 0, 1 and 2, we see that the structure which emerges is that of the arithmetic of addition modulo 3, i.e. a realization of the cyclic group of order 3,  C3  :

 + mod 3 0 1 2 period 0 0 1 2 1 1 1 2 0 3 2 2 0 1 3

Table 6.4

Table of C3
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Indeed we see that the global structure of C12, as it is depicted in Table 6.3, is mirrored in its homomorphic image C3 (But recall that it is not an isomorphism, since there is no 1,1 correspondence between the elements of the two groups.
In the following we summarize this homomorphism, which we call  m, of our group C12, a homomorphism with respect to the subgroup R = {0, 3, 6, 9}, resulting in an image group with the structure of C3 :

 + mod 12 0 3 6 9 1 4 7 10 2 5 8 11 0 0 3 6 9 1 4 7 10 2 5 8 11 3 3 6 9 0 4 7 10 1 5 8 11 2 6 6 9 0 3 7 10 1 4 8 11 2 5 9 9 0 3 6 10 1 4 7 11 2 5 8 1 1 4 7 10 2 5 8 11 3 6 9 0 4 4 7 10 1 5 8 11 2 6 9 0 3 7 7 10 1 4 8 11 2 5 9 0 3 6 10 10 1 4 7 11 2 5 8 0 3 6 9 2 2 5 8 11 3 6 9 0 4 7 10 1 5 5 8 11 2 6 9 0 3 7 10 1 4 8 8 11 2 5 9 0 3 6 10 1 4 7 11 11 2 5 8 0 3 6 9 1 4 7 10
 + mod 3 0 1 2 0 0 1 2 1 1 2 0 2 2 0 1
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Finally, we will discuss that homomorphism of C12 which is based on the subgroup S = {0, 2, 4, 6, 8, 10}.
The next group table (of C12) shows the cosets of the subgroup S (remember that each square in the table represents a coset, we must however count each element only once, i.e. if we find, say, two fours in such a square, we count it as one four). The elements of the group are arranged such that the subgroup S = {0, 2, 4, 6, 8, 10} stands out.

 + mod 12 0 2 4 6 8 10 1 3 5 7 9 11 period 0 0 2 4 6 8 10 1 3 5 7 9 11 1 2 2 4 6 8 10 0 3 5 7 9 11 1 6 4 4 6 8 10 0 2 5 7 9 11 1 3 3 6 6 8 10 0 2 4 7 9 11 1 3 5 2 8 8 10 0 2 4 6 9 11 1 3 5 7 3 10 10 0 2 4 6 8 11 1 3 5 7 9 6 1 1 3 5 7 9 11 2 4 6 8 10 0 12 3 3 5 7 9 11 1 4 6 8 10 0 2 4 5 5 7 9 11 1 3 6 8 10 0 2 4 12 7 7 9 11 1 3 5 8 10 0 2 4 6 12 9 9 11 1 3 5 7 10 0 2 4 6 8 4 11 11 1 3 5 7 9 0 2 4 6 8 10 12

Table 6.5

Table of C12   (version e).
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In the above table we see that the two cosets of the subgroup S = {0, 2, 4, 6, 8, 10} are the sets {0, 2, 4, 6, 8, 10} and {1, 3, 5, 7, 9, 11}. for every element of the group the left and right cosets are equal, i.e. for any element  x  of our group G the following relation holds :   xS = Sx. This implies that the subgroup S is a normal subgroup in G  (G = C12, realized as {0, 1, 2, 3, ..., 11}, + mod 12).

When we look to these two cosets we see that the one consists of EVEN numbers only, while the other consists of ODD numbers only. So we can represent them by this feature, i.e. we can set up the following homomorphism of G, with respect to the subgroup S :

{0, 2, 4, 6, 8, 10} -- EVEN
{1, 3, 5, 7, 9, 11} -- ODD

Let's call this homomorphism  q.

If we now in the above table replace the subset {0, 2, 4, 6, 8, 10} by EVEN, and the subset {1, 3, 5, 7, 11} by ODD, then we see that the set {EVEN, ODD} forms a group with the structure of C2, i.e. the cyclic group of order 2  :

 + EVEN ODD period EVEN EVEN ODD 1 ODD ODD EVEN 2

Table 6.6

Table of C2, version A.
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In this group the group operation is addition (+) :
(an) EVEN (number) + (an) EVEN (number) is equal to (an) EVEN (number).
(an) ODD (number) + (an) ODD (number) is equal to (an) EVEN (number).
(an) ODD (number) + (an) EVEN (number) is equal to (an) ODD (number).

It is clear that this operation is commutative.
The first and third additions just shown indicate that EVEN is the identity element.

This group is of course isomorphic (i.e. fully equivalent) to the group (0, 1}, + mod 2  :

 + mod 2 0 1 period 0 0 1 1 1 1 0 2

Table 6.7

Table of C2, version B.
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For easy comparison we summarize the homomorphism  q of the group G with respect to the subgroup S = {0, 2, 4, 6, 8, 10} :

 + mod 12 0 2 4 6 8 10 1 3 5 7 9 11 0 0 2 4 6 8 10 1 3 5 7 9 11 2 2 4 6 8 10 0 3 5 7 9 11 1 4 4 6 8 10 0 2 5 7 9 11 1 3 6 6 8 10 0 2 4 7 9 11 1 3 5 8 8 10 0 2 4 6 9 11 1 3 5 7 10 10 0 2 4 6 8 11 1 3 5 7 9 1 1 3 5 7 9 11 2 4 6 8 10 0 3 3 5 7 9 11 1 4 6 8 10 0 2 5 5 7 9 11 1 3 6 8 10 0 2 4 7 7 9 11 1 3 5 8 10 0 2 4 6 9 9 11 1 3 5 7 10 0 2 4 6 8 11 11 1 3 5 7 9 0 2 4 6 8 10
 + mod 2 0 1 0 0 1 1 1 0
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This concludes our exposition about the homomorphisms of the group C12 supported by its several subgroups.
In the next document we will continue our study of cyclic groups of composite order.
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To continue click HERE for further group theoretic preparation to the study of the structure of three-dimensional crystals
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